Question: The matrices
\[\begin{pmatrix} a & 1 & b \\ 2 & 2 & 3 \\ c & 5 & d \end{pmatrix} \quad \text{and} \quad \begin{pmatrix} -5 & e & -11 \\ f & -13 & g \\ 2 & h & 4 \end{pmatrix}\]are inverses.  Find $a + b + c + d + e + f + g + h.$
Answer: The product of the matrices is
\[\begin{pmatrix} a & 1 & b \\ 2 & 2 & 3 \\ c & 5 & d \end{pmatrix} \begin{pmatrix} -5 & e & -11 \\ f & -13 & g \\ 2 & h & 4 \end{pmatrix} = \begin{pmatrix} -5a + f + 2b & ae - 13 + bh & -11a + g + 4b \\ -10 + 2f + 6 & 2e - 26 + 3h & -22 + 2g + 12 \\ -5c + 5f + 2d & ce - 65 + dh & -11c + 5g + 4d \end{pmatrix}.\]We have that $-10 + 2f + 6 = -22 + 2g + 12 = 0,$ so $f = 2$ and $g = 5.$

Then
\[\begin{pmatrix} a & 1 & b \\ 2 & 2 & 3 \\ c & 5 & d \end{pmatrix} \begin{pmatrix} -5 & e & -11 \\ 2 & -13 & 5 \\ 2 & h & 4 \end{pmatrix} = \begin{pmatrix} -5a + 2 + 2b & ae - 13 + bh & -11a + 5 + 4b \\ 0 & 2e - 26 + 3h & 0 \\ -5c + 10 + 2d & ce - 65 + dh & -11c + 25 + 4d \end{pmatrix}.\]This gives us $-5a + 2 + 2b = 1,$ $-11a + 5 + 4b = 0,$ $-5c + 10 + 2d = 0,$ and $-11c + 25 + 4d = 1.$  Solving these equations, we find $a = 3,$ $b = 7,$ $c = 4,$ and $d = 5.$

Hence, $3e - 13 + 7h = 0,$ $2e - 26 + 3h = 1,$ and $4e - 65 + 5h = 0.$  Solving, we find $e = 30$ and $h = -11.$

Therefore, $a + b + c + d + e + f + g + h = 3 + 7 + 4 + 5 + 30 + 2 + 5 + (-11) = \boxed{45}.$